Joining two solids changes which faces are visible. Their volumes can be added when their interiors do not overlap, but their complete surface areas cannot simply be added: the touching faces become internal. A useful solution therefore needs two separate accounts, one for occupied space and one for exposed surfaces.
In this lesson, the components meet exactly along matching faces. We calculate a cylinder joined to a hemisphere and a cuboid joined to a triangular prism, then adapt the method to an open container. No density, flotation, material loss or frustum formula is needed. Dimensions are supplied rather than measured from the illustrations.
Make a face list before using formulas
For volume, identify the component dimensions and calculate each component once. A shared boundary has no thickness and adds no volume, so there is no join-volume term to subtract in these examples. If two components overlapped through a three-dimensional region, simple addition would count that region twice; that different situation is not used here.
For exposed area, imagine tracing the outside with a finger. Count each external face once. A face completely covered by the other solid contributes zero. If you start by adding both components’ closed surface areas, subtract the join area twice, because it appeared once in each component’s total.
| Component | Volume | Relevant surface area |
|---|---|---|
| Cylinder of radius r and height h | πr2h | Curved area 2πrh; each circular end πr2 |
| Hemisphere of radius r | (2/3)πr3 | Curved area 2πr2; flat circular face πr2 |
| Cuboid of length l, width w, height h | lwh | Closed area 2(lw + lh + wh) |
| Right triangular prism of length l | Triangle area × l | Two triangular ends plus one rectangle for each triangle side |
A hemisphere’s curved area is half a sphere’s surface area. Its closed surface area also includes the flat circle, giving 3πr2. Keep these two meanings separate: at a circular join, the flat circle disappears from the exposed-area calculation.
Worked case 1: a closed cylinder with one hemispherical cap
A solid consists of a cylinder of radius 3 cm and height 8 cm, joined to one hemisphere of radius 3 cm. The hemisphere’s flat face completely covers the cylinder’s top circle. The cylinder’s bottom circle remains exposed. Find the total volume, exposed area and total height.
Exploded-view data: the hemisphere and cylinder have the same 3 cm radius. Their matching circular faces are shown separated only to reveal the join; they coincide in the finished solid. The cylinder height is 8 cm, the hemisphere adds 3 cm vertically, and the lower cylinder disk remains an outside face.
Cylinder volume = π × 32 × 8 = 72π cm3. Hemisphere volume = (2/3)π × 33 = 18π cm3. Total volume = 90π cm3 ≈ 282.74 cm3, using calculator π.
| Surface | Counted area | Reason |
|---|---|---|
| Cylinder curved wall | 2π × 3 × 8 = 48π cm2 | Outside surface |
| Hemisphere curved cap | 2π × 32 = 18π cm2 | Outside surface |
| Cylinder bottom circle | π × 32 = 9π cm2 | Exposed flat base |
| Two matching join faces | 0 cm2 | Internal in the assembled solid |
The exposed area is 48π + 18π + 9π = 75π cm2 ≈ 235.62 cm2. As an independent check, closed component areas total 66π + 27π = 93π cm2. Subtract the two 9π join faces: 93π − 18π = 75π cm2.
The total height is 8 + 3 = 11 cm. Using 11 cm as the cylinder height would add a cylinder-shaped volume where the actual cap curves inwards. Read a height label as belonging to its named component, not automatically to the whole object.
Worked case 2: a cuboid with a triangular-prism roof
A closed model consists of a cuboid 10 cm long, 6 cm wide and 4 cm high, with a right triangular prism attached along the entire 10 cm by 6 cm top face. The prism has length 10 cm and an isosceles triangular cross-section with base 6 cm and perpendicular height 4 cm. Find the volume and exposed surface area.
Exploded-view data: the cuboid’s top and the prism’s rectangular base both measure 10 cm by 6 cm. The roof triangle has height 4 cm and half-base 3 cm, so each sloping side is 5 cm. The gap in the illustration only reveals the join. The roof and cuboid meet in the assembled model.
The triangular cross-section has area (1/2) × 6 × 4 = 12 cm2. Its sloping side is √(32 + 42) = 5 cm. The cuboid volume is 10 × 6 × 4 = 240 cm3, while the prism volume is 12 × 10 = 120 cm3. Total volume = 360 cm3.
Count the exposed cuboid faces: bottom 10 × 6 = 60 cm2; two long walls 2 × 10 × 4 = 80 cm2; two short ends 2 × 6 × 4 = 48 cm2. Their subtotal is 188 cm2. The top is entirely covered, so do not add its 60 cm2.
The roof contributes two sloping rectangles, 2 × 10 × 5 = 100 cm2, and two triangular ends, 2 × 12 = 24 cm2. Its exposed subtotal is 124 cm2. The complete exposed area is 188 + 124 = 312 cm2.
Check by subtraction: the closed cuboid area is 248 cm2; the closed prism area is 2 × 12 + 10(6 + 5 + 5) = 184 cm2. Subtract both copies of the 60 cm2 join: 248 + 184 − 120 = 312 cm2. The roof height of 4 cm belongs in its triangular area, while its sloping length of 5 cm belongs in the roof rectangles.
Worked case 3: an open container with a rounded bottom
An ideal thin-walled container has a vertical cylindrical section of internal radius 4 cm and height 9 cm above a hemispherical bowl of internal radius 4 cm. The top is open. Find its capacity and the internal curved area that needs coating. Ignore wall thickness and rim area; coat only the inside, not both sides of the material.
Capacity is cylinder volume plus hemisphere volume: π × 42 × 9 + (2/3)π × 43 = 144π + 128π/3 = 560π/3 cm3 ≈ 586.43 cm3. The cylinder and bowl communicate through their shared circle, so no disk blocks the interior.
The coating area is the cylindrical wall plus the hemisphere’s curved interior: 2π × 4 × 9 + 2π × 42 = 72π + 32π = 104π cm2 ≈ 326.73 cm2. There is no top disk because the container is open, and there is no flat bottom disk because its bottom is hemispherical.
Convert volume units by cubing the length factor
Since 1 m = 100 cm, a cubic metre contains 100 × 100 × 100 = 1,000,000 cubic centimetres. Divide a volume in cm3 by 1,000,000 to express it in m3. For example, 360 cm3 = 0.000360 m3. Dividing by only 100 would apply a length conversion to a volume.
For capacity, 1 litre = 1000 cm3. The open container therefore holds 560π/3000 litres ≈ 0.58643 litres. Converting its area would require a squared factor instead; surface area cannot be converted into litres.
Three practice tasks with full solutions
Practice 1: another closed capped cylinder
A cylinder of radius 2 cm and height 7 cm carries one matching hemisphere. Its other circular base is exposed. Find volume and exposed area. Volume = 28π + 16π/3 = 100π/3 cm3 ≈ 104.72 cm3. Area = 28π + 8π + 4π = 40π cm2 ≈ 125.66 cm2. The three area terms are cylinder wall, curved cap and bottom disk.
Practice 2: a wider roof model
A cuboid is 8 cm long, 8 cm wide and 3 cm high. A right triangular prism covers its whole top; the isosceles roof cross-section has base 8 cm and height 3 cm. Find volume and exposed area. Half-base 4 cm and height 3 cm give sloping sides of 5 cm.
Volume = 8 × 8 × 3 + [(1/2) × 8 × 3] × 8 = 192 + 96 = 288 cm3. Exposed cuboid area = 64 + 48 + 48 = 160 cm2. Roof area = 2 × 8 × 5 + 2 × 12 = 104 cm2. Total area = 264 cm2.
Practice 3: an open rounded-bottom vessel
A thin-walled open vessel has internal radius 3 cm, cylindrical height 5 cm and a matching hemispherical bottom. Find capacity in litres and its inside curved area, ignoring rim area. Capacity = 45π + 18π = 63π cm3 = 0.063π litres ≈ 0.19792 litres. Inside area = 30π + 18π = 48π cm2 ≈ 150.80 cm2. Adding a 9π top disk would incorrectly close the vessel.
Frequently asked questions
Why can I add volumes but not all surface areas?
The components occupy separate interior regions, while their touching faces become internal. Those shared faces have no exposed area.
Why subtract the join twice?
Only when starting from both closed component areas: each total already contains one copy of the matching join face.
Does every hemisphere contribute 3πr²?
No. That is its closed area including the flat circle. Its curved surface alone contributes 2πr².
Is cylinder height the total composite height?
No. With one hemispherical cap, total height is cylinder height plus the radius. Use each dimension for its own component.
Does an open container have a top area?
An open aperture has no material disk to coat. Include a rim only if its dimensions and the requested scope require it.
Should inside coating include outside surfaces?
No. Follow the named surface. These ideal-container tasks ask for the inside only and ignore material thickness.
How do I check the units?
Area uses squared units and volume uses cubed units. Cube length-conversion factors for volume; use 1000 cm³ per litre for capacity.
Build your own face account
List the components, matching joins and remaining outside faces before calculating. The Mathematics topical revision collection offers further navigation, with some contents potentially requiring membership. No specific solids worksheet or download is promised here.
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